CEMC Fermat, 2022

Since $\triangle PQR$ is
right-angled at $R$, then by the
Pythagorean Theorem, $$PQ^2 = PR^2 + QR^2 = 12^2 + 16^2 = 144 + 256 = 400$$ Since $\text{PQ gt;0}$, then $PQ = 20$.

Since $M$ is the midpoint of $PQ$, then $MQ = \frac{1}{2}PQ = 10$.

Now $\triangle NMQ$ is similar to
$\triangle PRQ$, since each is
right-angled and they share a common angle at $Q$.

Therefore, $\dfrac{NQ}{PQ} = \dfrac{MQ}{RQ}$ and so $\dfrac{NQ}{20} = \dfrac{10}{16}$ which
gives $NQ = \dfrac{20 \cdot 10}{16} = \dfrac{25}{2}$.

Thus, $RN = RQ - NQ = 16 - \dfrac{25}{2} = \dfrac{32}{2} - \dfrac{25}{2} = \dfrac{7}{2}$.

Since $\triangle PNR$ is right-angled
at $R$, its area equals $\dfrac{1}{2} \cdot PR \cdot RN = \dfrac{1}{2} \cdot 12 \cdot \dfrac{7}{2} = 21$.